Question

The calculation of concentrations of the molecular and ionic species in an aqueous solution that has an analytical composition of 0.010 M KH2PO4 + 0.060 M K2HPO4. NOTE: The Ka's for the three acidic protons of phosphoric acid H3PO4 are:

1) 7.11 x 10-3

2) 6.32 x 10-8

3) 4.5 x 10-13.

It may be helpful to write out the three dissociation reactions along with their Ka's and pKa's before starting these problems.

What is the pH of this solution?

7.98

K2 for H3PO4 is the appropriate Ka to use here. Set up an equil. reaction of H2PO4- going to HPO4-2 + H+.

*So sorry for the grammer, but I am wondering why the second dissociation constant is the relevant one here. How do you decide this? Please give a detailed explination if possible, Thank You!

Answer 1

when we have weak acid and its conjugate base we say we have a buffer

here H2PO4- ( or KH2PO4) is weak acid and its conjugate base is HPO4^2- ( K2HPO4)

Ka of KH2PO4 = 6.32 x 10^-8 , pka = -log Ka = -log ( 6.32 x 10^-8) = 7.2

pH of buffer = pka of acid + log [ conjugate base] / [ acid]

        = 7.2 + log ( 0.06/0.01)

         = 7.98

or else we can do in other manner

H2PO4^-(aq)   <---> HPO4^2-(aq) + H+ (aq)

Ka = [HPO4^2-] [H+] /[H2PO4-]

6.3 x 10^-8 = (0.06) [H+] / ( 0.01)

we find [H+] and thus pH which will yield again 7.98